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12:00 AM
so what you're doing is this: you take a generator for each generator of $A$ (of course you can choose $S_A=A$ if you want) and a generator for each generator of $B$ and the only relations we're enforcing are the relations coming from $A$ and $B$ and we want the elements of $C$ to become trivial in the amalgamated product
so we're glueing together $A$ and $B$ along the common subgroup $C$
you have canonical homomorphisms $A \to A \ast_C B$ and $B \to A\ast_C B$
 
Hmmm, alright, let me write something up and post it in here
 
Is there a fast way to check if 2 is a generator of large prime ?
I think what am asking is , is there some criteria on the prime p such that 2 is a generator of F*_p
 
It's unknown (according to experts who told me when I asked) how to predict who generates as a function of $p$.
 
haha bad luck for me then
 
If there were $a_3\in A_3$ such that $a_3\notin\text{Im}(f_2)$ then let $\text{Im}(f_2)=\langle F\mid R_1\rangle$ and $\langle a_3\rangle=\langle a_3\vert R_2\rangle$ where $F$ and $\{a_3\}$ are sets of generators and $R_1,R_2$ are sets of relations. Then let $D=\langle F, a_3\mid R_1\cup R_2\cup(\langle a_3\rangle\cap\text{Im}(f_2))\rangle$
 
12:07 AM
@TedShifrin Why did you ask for that question ?
just curious
 
Is that a correct way to describe the amalgamated free product?
 
Because I wrote an exercise on the group theory of card shuffling and wanted to know a general formula for how many perfect shuffles it takes to return the deck to its original order.
 
Neat :)
 
That's precisely asking for the order of $2$ in the multiplicative group.
 
In number theory, Artin's conjecture on primitive roots states that a given integer a that is neither a perfect square nor −1 is a primitive root modulo infinitely many primes p. The conjecture also ascribes an asymptotic density to these primes. This conjectural density equals Artin's constant or a rational multiple thereof. The conjecture was made by Emil Artin to Helmut Hasse on September 27, 1927, according to the latter's diary. The conjecture is still unresolved as of 2019. In fact, there is no single value of a for which Artin's conjecture is proved. == Formulation == Let a be an integer...
Under GRH, 2 is a generator of 37.4% of the primes
 
12:16 AM
It seems like most theorems about primes are probalistic
 
because they're related to distributions
distribution <=> measure theory <=> probability
 
the other theorems are unsolved
what I mean is , there is too much we dont know yet about primes
 
12:40 AM
@Jacksoja for some philosophical remarks on that, see the first few paragraphs of this Terry Tao blog post: terrytao.wordpress.com/2015/01/04/…
 
@Semiclassical thanks
 
12:54 AM
hey @Daminark
 
In general topological spaces
does the set of limit points of a set $A$
contain its limit points?
 
@topologicalmagician I think so
 
1:17 AM
hey chat
I just saw a question about matrix multiplication and that lead me to another question: the ring $\mathcal M_n(F)$ of $n\times n$ matrices over a field $F$ can be seen as the ring of linear transformations from $F^n$ to itself. Is there any analogy with modules to talk about this when it's $\mathcal M_n(R)$, where $R$ is a commutative ring?
 
@LucasHenrique yes
just replace $F^n$ with the module $M^n$
 
$R^n$
 
yeah, that
 
I suppose that most of the theorems from linear algebra won't work, though
 
you still have Cayley--Hamilton
 
1:28 AM
things get a lot more subtle, yes
for example, for vector spaces, there's one notion of finiteness: finite dimension. For modules you have a whole zoo of finiteness conditions: finitely generated, finitely presented, coherent, finitely cogenerated, Noetherian, Artinian, finite length etc.
for vector spaces, these all agree
 
1:49 AM
Note to self: Chat doesn't have xymatrix enabled
I've been trying to prove exactness in this one sequence for, like, 3 hours now. I've gotten everything except the fact that $\text{ker}(f_2)\subseteq\text{Im}(f_1)$
(That's always the most difficult bit)
I am given that $0\to\text{Hom}_R(A_3,D)\to\text{Hom}_R(A_2,D)\to\text{Hom}_R(A_1,D)$ is exact for all $D$
Where the homomorphisms are the induced homomorphisms from $f_1$ and $f_2$. Ie $g_1(\psi)=\psi\circ f_1$ where $g_1:\text{Hom}_R(A_2,D)\to\text{Hom}_R(A_1,D)$ and $\psi\in\text{Hom}_R(A_2,D)$
 
@LeakyNun it only holds for $T_1$ spaces
Let $(X,d)$ be a metric space.

Show that if $q\in B(x,r)$ and $q\neq x$ then there exists $0<R<d(p,q)$ such that $B(q,R)\subseteq B(x,r)$.
 
@LukasHeger I don't want to bother you, but do you have any insight into the above problem? I've reached the point of being tired of beating my head against this particular wall
 
@Rithaniel what do you want to prove?
 
That $\text{ker}(f_2)\subseteq\text{Im}(f_1)$
Well, more generally, that $A_1\to A_2\to A_3\to 0$ is an exact sequence
where $f_i:A_i\to A_{i+1}$
 
the Yoneda functor $R-Mod \to [R-Mod^{op},Ab]$ is faithful and additive and hence it reflects exactness
 
2:01 AM
I suspect this might be a bit beyond my knowledge
 
How do I prove this:
Let $(X,d)$ be a metric space.

Show that if $q\in B(x,r)$ and $q\neq x$ then there exists $0<R<d(q,x)$ such that $B(q,R)\subseteq B(x,r)$.
I mean if $B(q,r')\subseteq B(x,r)$ it might not be the case that $r'\leq r$
 
Well, thank you for the help, Lukas
 
@Rithaniel @LukasHeger any ideas?
 
My brain is a mess right now, so I might not be the best person to ask. I am not fully understanding what I read right now
 
Whaddup
@lukas how are you doing?
 
2:19 AM
guys which book shows set theory $\cap$ and $\cup$ from logic operators like $\wedge$ , $\vee$ etc.
Its hard to find such a book.
 
Heya Daminark
 
How's it going Rithaniel?
 
I have a headache trying to figure out a category theory problem using what I know from algebra
Also, I'm eating grapes. How about you?
 
Trying to read through some Silverman
He might get better later but his makeshift intro to varieties and algebraic curves is... definitely makeshift
 
Might be that he has his own understanding of these things and describes it as he understands them
(I've realized that my approach to this problem isn't going to get me anywhere, I think)
 
2:33 AM
It's not quite that, more that he's trying to hit that "just enough" spot
 
Ah, so he's trying not to spend too much time on the introduction
 
You kinda need some algebraic geometry setup for elliptic curves but not that that much
In general you can define an elliptic curve as a smooth projective curve of genus 1 with a specified basepoint, and then prove that you can change variables and put such a curve in Weierstrass form
Also it ends up being noteworthy that a certain group attached to general varieties happens to be isomorphic to the elliptic curve as a group
But anyway my impression is that once you sorta get through that part, you just work with the Weierstrass equation and things are more explicit
 
@Rithaniel take $D=A_2/\mathrm{Im}(f_1)$
then the projection $\pi:A_2 \to D$ is in the kernel of $g_1$
as by construction $0=\pi \circ f_1=g_1(\pi)$
so by exactness we find $\psi \in \mathrm{Hom}_R(A_3,D)$ such that $\psi \circ f_2 = \pi$
 
Huh, I think I see it. I was trying to come up with an appropriate $D$ but thought I'd keep ending up with dead ends. But moving the $f_1$ into the $D$ might change things
Oooooooh, got it!
Thank you again, Lukas
 
2:48 AM
@Daminark you already worked with elliptic curves, right?
for the $(a+bi)(c+id)$ thing
 
Yeah, specifically I was gunning for Kronecker-Weber for $\mathbb{Q}(i)$ using $y^2 = x^3 - x$
 
nice
 
I got through the proof that you always get an abelian extension when you adjoin torsion points of that guy to $\mathbb{Q}(i)$, started trying the other direction but didn't make it all too far
 
yeah I imagine it's not easy
 
I might revisit it soon, I think I still have the document
 
3:35 AM
Well, howdy, Demonark @Daminark
 
Hey Ted, how's it going?
 
 
2 hours later…
5:35 AM
5
Q: Was this spherical device used in weather forecasting in the 1960's?

uhohThe New York Times obituary Robert M. White, Who Revolutionized Weather Forecasts, Dies at 92 shows the image below. It looks like it might be a model of the Earth but I can't tell for sure. Can the sphere be identified, and if so, is it related to weather forecasting? click for full size ...

The image in the question reminds me of the term "projective geometry". Any thoughts?
 
 
1 hour later…
6:40 AM
@anakhro I think I have fixed the mistake now - a sequence of points in the open neighborhood's points there isn't a convergent because it is supposed to be an infinite sequence.
 
 
2 hours later…
8:42 AM
@TedShifrin Hi Ted! Thanks for your message. I had also found Fulton's book:D (good to know you're recommending it too).
 
@Lukas Are you around by any chance? I'm having an hard time understanding what (co)products look like in a subobject category
 
 
2 hours later…
10:17 AM
Is $\prod\limits_{i=1}^\infty\mathbb{Z}$ isomorphic to any more other less-difficult-to-type-out groups?
 
10:36 AM
The entries in the sequence of the partial sums of the Möbius inverse of the Harmonic numbers is a subset of the inverse error function when put in the context of tuples in combinatorics.
 
10:54 AM
Hello
 
@Rithaniel I guess not
@AlessandroCodenotti what is a subobject category?
 
Just to double check, division and multiplication have the same level of precedence, correct? It's not division is before multiplication?
 
right
 
BEDMAS confused me...I thought since D comes before M it meant division is first :\
 
11:57 AM
That's a good video thanks
 
@LeakyNun You have a category $C$ and an object $X\in C$. The category of subobjects of $X$ has as objects objects $A$ of $C$ that have a mono $A\to X$, up to equivalence
 
If the span of an empty set is equal to the zero vector, what is the span of the subset containing the zero vector.
Is it also, the zero vector?
 
@Alessandro I know what coproducts look like if $C$ is abelian
 
Well that's not quite accurate, you really want monomorphisms into $X$ up to equivalence
 
not sure in general
 
12:08 PM
@LukasHeger I'm trying to understand why this categorical presentation of a topology actually is a topology mathoverflow.net/questions/265522/…
But I don't quite see what product and coproducts in the subobject category are
@MadSpaceMemer yes
 
okay so if you have two subobjects $i:A \to X$ and $j:B \to X$, then we can take the pullback $A \times_X B \to X$, monos are closed under composition and pullbacks (exercise) so that's actually a subobject
 
I guess you just get unions and intersections here?
 
if you do this in the category of sets (or groups or topological spaces) you do get intersections yeah
I think for coproducts it's more complicated
 
I mean, you do get unions for coproducts in Set
 
12:12 PM
The subobject category of $X$ is a subcategory of the slice category of $X$ right?
 
yes
 
So products in this category are just fibered products in the original category
 
xwhich is what I said
but you need to make sure that you do get a monomorphism
 
@LukasHeger Yes but I know from AG that this works out :P
 
but the general construction for coproducts requires some form of epi-mono factorization I think
take the coproduct $A \sqcup B \to X$, that won't be a monomorphism
 
12:14 PM
Right
 
but if $\mathcal C$ admits canonical epi-mono factorization, then we can factor it as a composition of epi by mono
so just take the mono part
in Set, that gives you unions
in Ab, that gives you sum of subgroups
etc.
 
Makes sense
Thanks
 
more generally, if $\mathcal C$ admits images in the sense of category theory, then we can take the image of $A \sqcup B \to X$
see
In category theory, a branch of mathematics, the image of a morphism is a generalization of the image of a function. == General Definition == Given a category C {\displaystyle C} and a morphism f : X → Y {\displaystyle f\colon X\to Y} in C {\displaystyle C} , the image of f {\displaystyle f} is a monomorphism m : I → Y ...
 
I mean products are just "intersections" right
 
yes
 
12:18 PM
@AlessandroCodenotti play?
 
@LeakyNun but intersections are just fiber products along monomorphisms, right?
 
@LukasHeger you're not wrong
 
12:39 PM
@LukasHeger should the plural of "anno domini" be "annis domini"?
 
yes
it's ablative
 
1:02 PM
@LeakyNun I wasn't there, sorry
 
now?
 
Just a game or two, I have to study
 
Henlo
 
Hi @Edward
 
What's up? :)
 
1:03 PM
Send me a challenge @Leaky
@EdwardEvans Not much, just reading some cool set theory, what about you?
 
Just reading ANT notes lol
In the UK
 
Nice
I'm also back to Italy
 
Nice :D
 
Hi @Edward
 
Hey @Lukas
 
1:14 PM
Hey, I have operation
1 / 1000 / 60 / 60 / 24

how to make it shorter?
i.e. making shorter the
1 / 2 / 2

would be:
1 / 4

I am performing multiple divisions and seek to perform just one
 
@Lukas I emailed Studiumsverwaltung about the Semestergebühren lol, the bureaucrats at the university must thing I'm the most disorganised person alive
 
@EdwardEvans tbf you should have gotten an email one month prior to the deadline
the time span to pay your fees was 15.01.-15.02.
 
Yep :P Unfortunately I didn't know I was obligated to check my university email address since everything else is registered via my personal email address (this is common in the UK too!)
and I'm a disorganised person anyway rofl
 
Hmm sorry I thought its math chat not uni chat cya
 
lol, we can chat about whatever we want
 
1:20 PM
(Basically I'm a moron)
 
@AlessandroCodenotti ggs
 
gg @Leaky
You had a chance to capture my passed pawn in the first game when I first offered a rook trade
 
@AlessandroCodenotti as a logician, are you interested in categorical logic at all?
I think it seems cool
 
I don't even know what that is
 
@AlessandroCodenotti I'm sure we missed a lot of things in both games
 
1:22 PM
I'm not really interested in logic to be fair (as in studying modal/infinitary/other weird logics), just in set theory
@LeakyNun Stockfish says I missed mate in 10 in the second game lmao
And that even ignoring that I had a big advantage that I wasted, the position was equal when I won on time
 
oh and there were back rank issues which means I couldn't capture your passed pawn
 
@Alessandro the idea of categorical logic, as I understand is, is to rephrase logical concepts in categorical terms and that allows you to apply logical reasoning to give really intuitive proofs for some things in certain categories. You can also use categorical algebra to prove logic things, but I know even less about that
 
@LeakyNun Oh right, I didn't even notice. But I had the right bishop to attack the pinned knight on the back rank
 
the "internal logic of a category" is in general really weird though
 
@LukasHeger Uh, weird, I know nothing about any of this
Is this related to the elementary topos things?
 
1:26 PM
yes, absolutely
 
@AlessandroCodenotti exactly
 
I see
I want to learn some topos theory from a logician point of view eventually
 
@AlessandroCodenotti there's the book "Topoi - A categorical analysis of logic"
we could read it together, I'd be down
 
Because I've seen that there's some super weird generalization of forcing over topoi
@LukasHeger Who is the author? I can't find it on the usual Russian website
 
Goldblatt
it's the categorical analysis not a
I misremembered
So if $X$ is a reduced scheme and $\mathcal F$ is a quasi-coherent sheaf of finite type over $X$, then there is a dense open subset $U$ of $X$ such that $\mathcal F|_{U}$ is locally free
that's a hard exercise in algebraic geometry
but using categorical logic, you can "trivialize" it
 
1:32 PM
Hello, if we have rs dividing , x ( x^k - 1 ) , rs are primes
can we say that either one of x and x^k -1 divides rs?
 
in the internal logic of the Zariski topos associated to $X$, the statement corresponds to the fact that "every finitely generated vector space does not not possess a basis"
which is a theorem in intuitionistic logic
 
is there categorical logic?
 
yes
 
logic was not hard enough ??
haha
Logic is mathematic of mathematics
I do not want to know what categorical logic would be like
 
$\mathcal F$ being a quasi-coherent sheaf of finite type over a reduced scheme just translates to a vector space over a field in the internal logic of the Zariski topos
"not not" translates to "on a dense open subset"
not that for example "$\lnot \lnot \lnot \lnot P \Rightarrow \lnot \lnot P$ is a theorem of intuitionistic logic
 
1:35 PM
@LukasHeger Looks interesting, but I already have a lot of stuff to work on
My thesis, an exam in March, I'm reading a set theory book...
 
which means that if something holds on a dense open subset of a dense open subset, it holds on a dense open susbet
if interpreted in the Zariski topos
 
@LukasHeger do you have a particular book about categorial logic in mind?
You made me curious about the topic
 
@Jacksoja there's Goldblatt "Topoi - The Categorical Analysis of Logic"
 
lemme check it!
 
2:07 PM
@StupidQuestionsInc I’m enjoying that book very much, it’s a book I was searching for long time. Thank you for introducing me with that book.
 
3:04 PM
@Lukas Is there a way to see an HNN extension as a pushout in Grp? I think there should be since you can view gluing a space to itself along subspaces as a pushout in Top
 
@Alessandro I'm not sure
I don't think you can
even semidirect products don't admit a nice description as a colimit in Grp
they're 2-colimits though
probably a HNN extension is some 2-colimit, too
 
Hmm so is there some other categorical construction which becomes an HNN extension in Grp and a "gluing a space to itself along two subspaces" in Top? I'm just wondering because they seem like very close analogues so maybe they're two instances of a general construction
 
the problem is that the defining relation of an HNN extension uses elements
since you want the isomorphism to be given by conjugation
 
Right
I see why category theory doesn't like that
 
well, 1-category theory doesn't
consider Grp as a subcategory of Cat, the category of small categories
the latter comes with a notion of natural transformations
that's why it's naturally a 2-category
 
3:11 PM
@LukasHeger I know very little about that, but I'm following so far
 
if you restrict that structure to Grp, you can ask yourself what a natural transformation between two group homorphisms $f_1,f_2:G \to H$ is
as it turns out, it's just an element $h \in H$ such that $hf_1(g)=f_2(g)h$ for all $h$
i.e. $hf_1h^{-1}=f_2$
this looks more adequately equipped to handle HNN extensions
so any group presentation trivially implies a universal property for that group
 
@LukasHeger Aha, this is looking promising
 
in the case of HNN extensions, if you have two subgroups $K,H$ of $G$ and an isomorphism $\alpha:H\to K$, the universal property of the HNN extension $W$ (what's the notation here) it should be this: let $i_1:H \to G$ and $i_2:K \to G$ be the inclusions, then homomorphisms $f:W \to T$ are in natural bijection to homomorphisms $f:G \to T$ and a natural transformation of $f \circ i_1 \to f \circ i_2$
I think this means that we have some form of 2-colimit
but I'm not sure
you should ask this on MO
(or I can)
if you do, be sure to reference this question:
27
Q: Are semi-direct products categorical (co)limits?

Makhalan DuffProducts, are very elementary forms of categorical limits. My question is whether in the category of groups, semi-direct products are categorical limits. As was pointed in: http://unapologetic.wordpress.com/2007/03/08/split-exact-sequences-and-semidirect-products/ Bourbaki (General Topology, Pr...

 
3:36 PM
Very interesting, thanks
 
4:26 PM
Hi @Balarka, how was the talk?
 
Good! Everyone enjoyed it
I printed pictures of the Morin-Francis eversion out and distributed it at the end of the talk
These pictures: 1, 2
 
Given that a matrix has eigenvalue 1,2 and 3, I am required to find out the determinant of $A^2 + A^T$
 
Nice
Cool pictures too
 
Yeah
 
4:53 PM
Congrats, a @Balarka! :)
Hi, demonic @Alessandro.
 
@BalarkaSen did you shout "Gromov is God"?
 
5:06 PM
@LeakyNun Unfortunately no
 
I need to update my triangulation talk
 
Does that make it a rectangulation talk?
 
Lmao
What's the abstract, @MikeMiller
 
@TedShifrin I have misconception about surface integral.
 
Consider subintervals of $[0,1]$ such that

$E_k$ which has $9^k$ intervals, each of length $10^{-k}$, with total gaps in between the intervals being greater than or equal to $10^{-k}$.

Let $F=\cap_{k=1}E_k$ be non-empty.

Suppose $0<\delta<1$ is such that $\frac{1}{10^{k+1}}\leq \delta <\frac{1}{10^k}$ and that any interval of diameter at most $\delta$ , may intersect 2 intervals. Further suppose each intersected interval contains point of the intersection, $F$, **it follows that the least number of sets in any $\delta$ cover of $F$ is greater than or equal to $\frac{9^k}{2}$.**
 
5:17 PM
What misconception is that, @adesh?
 
May someone elaborate on the bold?
the last sentence
please
 
@TedShifrin Let's consider a vector field $\mathbf A$, draw a closed surface $\sigma$ and let $\mathbf A$ is continuous in/on $\sigma$. Let $A_n$ represents the outward normal component at every point on the surface then the surface integral is defined as $$\oint_{\sigma} A_n~d\sigma$$
 
You don't need a closed surface to talk about that, by the way. I would write $\mathbf A\cdot\mathbf n$, but yes.
 
And now consider this figure
I thought that for every point inside and on the surface I got take the dot product with the outward normal but today someone told me that when we do surface integral we do dot product (or take the component along the outward normal) only for point on the surface not inside.
Have I made myself clear?
 
We're talking about things on the surface. $\mathbf n$ doesn't even make sense inside.
You're measuring how much of $\mathbf A$ is going across $\sigma$.
And remember my comment. This makes sense for any oriented surface — it needn't be closed and it needn't even have an "inside."
 
5:24 PM
MY GOD! You're awesome.
Please keep on explaining
I'm loving it
 
LOL, @adesh. You can watch my YouTube videos on surface integrals. There are lots of examples. (But they do use the language of differential forms, as well as the classical flux discussion.)
 
@TedShifrin Can you please call me "Demonic Adesh" ROFL
 
LOL, Alessandro earned that appellation.
 
Hi @TedShifrin
How are you?
 
5:26 PM
Hi, @topologicalmagician.
 
@TedShifrin do you have any thoughts on the problem I wrote above?
I did draw pictures this time
 
I don't have thoughts. It seems sloppily written. "may intersect 2 intervals"? What intervals?
 
@TedShifrin Good Night
 
@TedShifrin in $E_k$
 
Night, @adesh.
@topologicalmagician: The problem is incomplete. Do we know $E_{k+1}\subset E_k$?
 
5:31 PM
@TedShifrin yeah
 
You realize that you should post a complete and clear question if you're going to bug us?
 
@TedShifrin yeah, sorry. I gathered the info from a particular proof.
 
That doesn't help me.
 
Ok, the problem is to find the box dimension of the subset of $[0,1]$ which does not contain the digit 5 in their decimal expansion
 
What is box dimension?
 
5:35 PM
$lim_{\delta\rightarrow 0} \frac{N_d(F)}{-log d}$ where $N_d(F)$ is the least amount of sets of diameter at most d
@TedShifrin
 
Balls of diameter $\delta$?
This looks a lot like Hausdorff dimension.
 
@TedShifrin no, not balls. Just sets.
But I am able to compute part of it. The part I\m unable to understand is the following:
 
Presumably you get the smallest $N_\delta$ when you take an interval (or ball) in your case.
 
Consider subintervals of $[0,1]$ such that

$E_k$ which has $9^k$ intervals, each of length $10^{-k}$, with total gaps in between the intervals being greater than or equal to $10^{-k}$.
Note that $E_{k+1}\subseteq E_k$. for each k.
Let $F=\cap_{k=1}E_k$ be non-empty.

Suppose $0<\delta<1$ is such that $\frac{1}{10^{k+1}}\leq \delta <\frac{1}{10^k}$ and that any interval of diameter at most $\delta$ , may intersect 2 intervals in $E_k$. Further suppose each intersected interval contains point of the intersection, $F$, **it follows that the least number of sets in any $\delta$ cover of $F$ is
(I edited it)
Yeah, the only thing I'm unable to understand, is why $N_d(F)\geq \frac{9}{2}$
 
That dimension definition makes no sense to me. You mean $N_\delta\ge 9^k/2$?
 
5:43 PM
@TedShifrin yes, if by $N_\delta$ you mean the least amount of sets of diameter at most delta required to cover the set, $F$.
 
So what does that make the box dimension in this case?
 
log9\log10
but that's because $N_d(F)\leq 9^k$ for $0<\delta<1$ (This I'm able to prove)
 
Where did the /2 go?
 
@BalarkaSen What's a triangulation, what's the triangulation conjecture, high-D stuff reducible to homology cobordism in 3D, modern tools to attack hcob
 
@TedShifrin observe that for $0<\delta<1$, there exists a $k\in \mathbb{N}$ such that $\frac{1}{10^k}\leq \delta < \frac{1}{10}^{k-1}$ and so every interval of diameter at most delta is a delta cover for $F$, since there are $9^k$ such intervals, $N_{\delta}(F)\leq 9^k$
 
5:50 PM
I do not follow.
 
@TedShifrin we were told to only consider $0<\delta<1$
 
I'm not going to be able to help. My brain isn't working for this kind of stuff today.
 
@TedShifrin alright. No worries. I'll try to figure it out.
@TedShifrin oh! I got it
I mean I understood it
because he's removing intervals
the picture I just drew helped
@TedShifrin fractal geometry is interesting but annoying lol
 
@adeshmishra glad you found it useful! and best wishes for what comes next ;)
 
Cool @topologicalmagician
 
5:59 PM
hello, if $a\in [0,\pi]$ and $b\in [\pi/2,\pi]$ how to prove that cos(a)=cos(b) ?
 
That is nonsense, @Poline.
 
I want to find $a\in[0,\pi]$ such that $cos(a)=cos(b)$ and $b\in[\pi/2,\pi]$
 
Obviously, then, take $a=b$.
 
I don't see why a=b
that is my problem
 
Look at the graph.
 
6:02 PM
@PolineSandra if you want something that works then $a=b$ satisfies your conditions
also look at the graph as Ted suggeste
 
The point is that it's the only thing that works.
Cosine is one-to-one on $[0,\pi]$.
Then there's a unique $a\in [\pi/2,\pi]$ with $\cos a = \cos b$.
And you can figure it out from the unit circle picture.
 
@TedShifrin yes it is true I forget this
and if $b\in [\pi,3\pi/2]$ ?
 
Why did you erase this? I answered it already.
 
your answer is about $\pi/2,\pi]$
not about $[\pi,3\pi/2]$
 
No, it was about this question. You need to review basic trigonometry.
 
6:07 PM
@TedShifrin is it -b ?
@TedShifrin I'm sorry I don't read carefully
 
Where does $-b$ live if $b\in [\pi,3\pi/2]$?
 
$b\in [\pi,3\pi/2]$ but there is c symmetric to b in $[\pi/2,\pi]$
@TedShifrin
or $a=b-\pi$ @TedShifrin
 
6:22 PM
No. Stop making guesses and look at either the unit circle or the graph of the cosine function
What does "c symmetric to b" mean?
 
$c=-b\in [\pi/2,\pi]$
 
GROWL.
Come on. $-b\in [-3\pi/2,-\pi]$. We're talking numbers here, not points on the unit circle.
So you're close but you need to fix it.
 
if $b\in [\pi,3\pi/2]$ then $b-\pi\in[0,\pi]$
but $cos(a)\neq cos(b-\pi)$
 
You're back to writing symbols instead of thinking.
 
I draw the circle and c but I can see what is equal c
 
6:38 PM
How do you get c from b?
 
symmetry from the axe of abscise
 
OK, reflecting across the $x$-axis. So the angle $b\in [\pi,3\pi/2]$ goes to what angle when you do that?
 
[\pi/2,\pi]
 
Yes, but I want you to use the geometry to give me a correct formula.
 
I don't know how ?
 
6:48 PM
What is the angle from the negative $x$-axis to the ray at angle $b$?
 
I don't know if I understand exactly (I speak French) but it is $\pi+b$
 
Non, on commence au point $(-1,0)$ et on va jusqu'au point $(\cos b,\sin b)$.
C'est quel angle?
 
$3\pi/2-b$
 
Vraiment? On a commencé à l'angle $\pi$, n'est-ce pas?
 
on pardon j'ai vu (0,-1), du point pi jusqu'a b c'est b-\pi
 
6:56 PM
Oui, c'est ça. Et pour commencer à $(-1,0)$ et arriver à $c$, on fait quoi?
 
$\pi-b$
 
Non. Il faut penser à ce qu'on vient de discuter.
 
je me suis trompé de sense
c'est $3\pi/2+b$
non pardon
 
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